Layer of Insurance

Study guides: Exam 7, Exam 8, Exam GI-301

Layer of Insurance is the slice of each loss between an attachment point $a$ and an exhaustion point $b = a + l$ , written " $l$ xs $a$ ", together with the coverage that pays it (primary, excess or reinsurance). Its cost is the expected loss that falls inside that slice.

$X_{(a,b]} = \min(X, b) - \min(X, a)$

$E\left[X_{(a,b]}\right] = E[X \wedge b] - E[X \wedge a]$

$E[\text{layer loss}] = E[N]\left(E[X \wedge b] - E[X \wedge a]\right)$

$X$ is the ground-up severity, $N$ the claim count and $E[X \wedge u]$ the limited expected value. Equivalently, $E[X_{(a,b]}] = \int_a^b S(x)\,dx$ with $S = 1 - F$ — the horizontal strip of a Lee diagram between $a$ and $b$ .
Frequency and severity in the layer. $E[N]\,S(a)$ claims are expected to reach the layer. Each pays on average $\left(E[X \wedge b] - E[X \wedge a]\right)/S(a)$ . This is how Frequency and Severity distributions give the expected losses by layer.
Cost given the overall loss cost. Layer loss cost $= \text{overall loss cost} \times \dfrac{E[X \wedge b] - E[X \wedge a]}{E[X]}$ . In ILF form, it is the basic-limits loss cost $\times\,[\text{ILF}(b) - \text{ILF}(a)]$ . The bottom layer below a deductible is the LER.
Properties. Layers stack: the layers of a tower add up to the ground-up loss. Expected cost per dollar of limit falls as the layer rises (the consistency test). Upper layers grow faster than the ground-up trend under inflation, and they report and develop later. That is why unpaid claims must be estimated layer by layer rather than by scaling ground-up results — see Excess Insurance and Excess and Deductible Rating.

Expected Losses in a Layer from Frequency and Severity Claims arrive with a Poisson mean of $20$ per year. Severity is Pareto with $\alpha = 2$ and $\theta = \$100{,}000$ , so $E[X \wedge u] = \dfrac{\theta u}{\theta + u}$ and $S(x) = \left(\dfrac{\theta}{\theta + x}\right)^2$ . Find the expected annual loss in the layer $\$400{,}000$ xs $\$100{,}000$ , the expected number of claims reaching it, and the average payment per such claim.
Answer
 $\begin{align*} E[X \wedge 500\text{K}] &= \frac{100{,}000 \times 500{,}000}{600{,}000} \\ &= 83{,}333.33 \\[4pt] E[X \wedge 100\text{K}] &= \frac{100{,}000 \times 100{,}000}{200{,}000} \\ &= 50{,}000.00 \\[4pt] \text{Per claim} &= 83{,}333.33 - 50{,}000.00 \\ &= 33{,}333.33 \\[4pt] \text{Aggregate} &= 20 \times 33{,}333.33 \\ &= \$666{,}667 \end{align*}$ 
 $\begin{align*} \text{Claims in layer} &= 20 \times \left(\tfrac{100}{200}\right)^2 \\ &= 5 \\[4pt] \text{Severity in layer} &= \frac{33{,}333.33}{0.25} \\ &= \$133{,}333 \end{align*}$ 
The layer carries one-third of the expected ground-up loss ( $E[X] = \theta = \$100{,}000$ per claim), from a quarter of the claims.

Cost of a Layer Given the Overall Loss Cost A ground-up loss cost of $\$2{,}400$ per unit of exposure has been set. Ten representative claims, in $\$000$ , are $20, 40, 60, 80, 120, 150, 250, 400, 700, 1{,}180$ (total $3{,}000$ ). Price the layer $\$500{,}000$ xs $\$250{,}000$ .
Answer Cap each claim at $750$ and at $250$ , then difference the totals:
 $\begin{align*} \textstyle\sum \min(X, 750) &= 2{,}570 \\ \textstyle\sum \min(X, 250) &= 1{,}470 \\[4pt] \text{Layer share} &= \frac{2{,}570 - 1{,}470}{3{,}000} \\ &= 0.3667 \\[4pt] \text{Layer loss cost} &= 2{,}400 \times 0.3667 \\ &= \$880 \end{align*}$ 
The layer takes $36.7\%$ of the loss cost. That figure rests on only three claims ( $400$ , $700$ and $1{,}180$ ), so in practice the share would come from a fitted severity curve or industry ILFs rather than from this sample.